Let $X$be a Poisson random variable with a mean of$20$.

$\left(a\right)$Use the Markov inequality to obtain an upper bound$p=P(X\ge 26)$.

$\left(b\right)$Use the one-sided Chebyshev inequality to obtain an upper bound$p$.

$\left(c\right)$Use the Chernoff bound to obtain an upper bound$p$.

$\left(d\right)$Approximate $p$by making use of the central limit theorem.

$\left(e\right)$Determine $p$by running an appropriate program.

Let $X$be a Poisson random variable with a mean of$20$.

Suppose that $X$is a Poisson random variable with a parameter$\lambda$. It is given that $X$is a random variable with a mean of$20$. Since the expected value and variance of a Poisson random variable are both equal to its parameter$\lambda ,X$has mean $\mu =\lambda =20$and variance${\sigma }^2=\lambda =20$.

By Markov's inequality,

$p=P\{X\ge 26\}\le \frac{E\left[X\right]}{26}=\frac{20}{26}=.7692$

Using Corollary$5.1$, localid="1649901568148" $\underset{¯}{a=6}$we get:

$p=P\{X\ge \underset{=26}{\underbrace{20+6}}\}\le \frac{{\sigma }^2}{{\sigma }^2+6^2}=\frac{20}{20+36}=.\mathbf{3571}$

See Example$5d$. Since $X$is a Poisson random variable with parameterlocalid="1649901587221" $\lambda =20$using the result

$P\{X\ge i\}\le \frac{e^{-\lambda }(e\lambda {)}^i}{i^i}$

We get,

$p=P\{X\ge 26\}\le \frac{e^{-20}(20e{)}^{26}}{{26}^{26}}=.4398$

Using the central limit theorem we get:

$p=P\{X\ge 26\}\stackrel{\text{the continuity correction}}{=}P(X\ge 25.5)$

$=1-P\{X<25.5\}=1-P\left\{\frac{X-20}{\sqrt{20}}<\frac{25.5-20}{\sqrt{20}}\right\}\approx 1-\Phi (1.23)$

$\text{Table 5.1 (textbook, Chapter 5)}1-.8907=.\mathbf{1093}\text{.}$

Using software package by personal choice, we obtain:

$p=P\{X\ge 26\}=1-P\{X<26\}=1-\sum _{i=0}^{25}{e}^{-20}\frac{{20}^i}{i!}$

$=1-.887815=\mathbf{0}\mathbf{.}\mathbf{112185}$